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Motion of a particle

  • Thread starter chart2006
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1. Homework Statement

The two-dimensional motion of a particle is defined by the relationship [tex] r = \frac {1}{sin\theta - cos\theta} [/tex] and [tex] tan\theta = 1 + \frac {1}{t^2} [/tex], where [tex] r [/tex] and [tex] \theta [/tex] are expressed in meters and radians, respectively, and [tex] t [/tex] is expressed in seconds. Determine (a) the magnitudes of velocity and acceleration at any instant, (b) the radius of curvature of the path.


2. Homework Equations

[tex] r = \frac {1}{sin\theta - cos\theta} [/tex]


[tex] tan\theta = 1 + \frac {1}{t^2} [/tex]



3. The Attempt at a Solution

I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex] tan\theta [/tex] for [tex] \theta [/tex]. Once i've done that I figure I'd need to differentiate both [tex] r [/tex] and [tex] \theta [/tex] to find [tex] \dot{r}, \ddot{r}, \dot{\theta}, \ddot{\theta}[/tex].

I don't know if I'm on the correct route but any help would be appreciated. thanks!
 

gabbagabbahey

Homework Helper
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I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex] tan\theta [/tex] for [tex] \theta [/tex].
Tangent is not a one-to-one function, so that's a bad idea.

Instead, draw a picture!:smile: I think you can find expressions for [itex]\sin\theta[/itex] and [itex]\cos\theta[/itex] in terms of [itex]t[/itex] without actually solving for [itex]\theta[/itex] first.....think 'right triangle':wink:
 

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